设函数f(x)=cos(2x+π/3)+sin方x1)求函数f(x)的最大值和最小正周期2)设A、B、C为△ABC的三个内角,若cosB=1/3,f(C/2)=-1/4,且C为锐角,求sinA
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![设函数f(x)=cos(2x+π/3)+sin方x1)求函数f(x)的最大值和最小正周期2)设A、B、C为△ABC的三个内角,若cosB=1/3,f(C/2)=-1/4,且C为锐角,求sinA](/uploads/image/z/1063675-19-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dcos%282x%2B%CF%80%2F3%29%2Bsin%E6%96%B9x1%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F2%EF%BC%89%E8%AE%BEA%E3%80%81B%E3%80%81C%E4%B8%BA%E2%96%B3ABC%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92%2C%E8%8B%A5cosB%3D1%2F3%2Cf%EF%BC%88C%2F2%29%3D-1%2F4%2C%E4%B8%94C%E4%B8%BA%E9%94%90%E8%A7%92%2C%E6%B1%82sinA)
设函数f(x)=cos(2x+π/3)+sin方x1)求函数f(x)的最大值和最小正周期2)设A、B、C为△ABC的三个内角,若cosB=1/3,f(C/2)=-1/4,且C为锐角,求sinA
设函数f(x)=cos(2x+π/3)+sin方x
1)求函数f(x)的最大值和最小正周期
2)设A、B、C为△ABC的三个内角,若cosB=1/3,f(C/2)=-1/4,且C为锐角,求sinA
设函数f(x)=cos(2x+π/3)+sin方x1)求函数f(x)的最大值和最小正周期2)设A、B、C为△ABC的三个内角,若cosB=1/3,f(C/2)=-1/4,且C为锐角,求sinA
f(x)=cos(2x+π/3)+sin²x
=cos2xcosπ/3-sin2xsinπ/3+[1-cos(2x)]/2
=1/2cos2x-√3/2sin2x+1/2-1/2cos2x
=-√3/2sin2x+1/2
当sin2x=-1时,最大值=(1+√3)/2
最小正周期=2π/2=π
f(C/2)=-√3/2sinC+1/2=-1/4
sinC=√3/2
∵C为锐角
C=π/3
cosC=1/2
∵cosB=1/3
∴sinB=2√2/3
sinA=sin(π-B-C)
=sin(B+C)
=sinBcosC+cosBsinC
=2√2/3×1/2+1/3×√3/2
=√2/3+√3/6
设函数f(X)=cos(2x+π/3)+sin方x。设A、B、C为△ABC的三个内角,设函数f(X)=cos(2x+π/3)+sin x。A、B、C为△ABC的三个内角,若
f(x)=cos(2x+π/3)+sin方x=cox2xcosπ/3-sin2xsinπ/3+(1-cos2x)/2
=1/2-根3/2*sin2x
因为-1<=sin2x<=1
所以:f(x)=1/2-根3/2*sin2x
1/2-根3/2《=f(x)<=1/2+根3/2
最大值为1/2+根3/2
1/2-根3/2*sinC=-1/4
si...
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f(x)=cos(2x+π/3)+sin方x=cox2xcosπ/3-sin2xsinπ/3+(1-cos2x)/2
=1/2-根3/2*sin2x
因为-1<=sin2x<=1
所以:f(x)=1/2-根3/2*sin2x
1/2-根3/2《=f(x)<=1/2+根3/2
最大值为1/2+根3/2
1/2-根3/2*sinC=-1/4
sinC=根3/2
C=π/3
cosB=1/3
B=arccos1/3
A=π-π/3-arccos1/2
sinA=sin(π/3+arccos1/2)
=sinπ/3cos(arccos1/3)+sin(arccos1/3)cosπ/3
=根3/2*1/3+2根2/3*1/2
=(根3+2根2)/6
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f(x)=[(1/2)cos2x-(√3/2)sin2x]+(1/2)[1-cos2x]=-(√3/2)sin2x+(1/2)
1、最大值是√3+(1/2),最小正周期是π;
2、f(C/2)=-1/4,得:sinC=√3/2,则C=π/3,又cosB=1/3,则sinB=2√2/3
sinA=sin(2π/3-B)=(√3/2)cosB+(1/2)sinB=[√3+2√2]/6
1)f(x)=cos(2x+π/3)+sin方x=cos2xcosπ/3-sin2xsinπ/3+(1-cos2x)/2
=1/2cos2x-根3/2sin2x+1/2-1/2cos2x=1/2-根3/2sin2x
所以最大值为1/2+根3/2,最小正周期为π
2)f(C/2)=-1/4,即1/2-根3/2sinC=-1/4,所以sinC=根3/2,...
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1)f(x)=cos(2x+π/3)+sin方x=cos2xcosπ/3-sin2xsinπ/3+(1-cos2x)/2
=1/2cos2x-根3/2sin2x+1/2-1/2cos2x=1/2-根3/2sin2x
所以最大值为1/2+根3/2,最小正周期为π
2)f(C/2)=-1/4,即1/2-根3/2sinC=-1/4,所以sinC=根3/2,因为C为锐角,所以C=π/6
所以sinA=sin(π-π/6-B)=sin(5π/6-B)=sin5π/6cosB-cos5π/6sinB=(1/2)*(1/3)-(根3/2)*(2根2)/3
=(1-2根6)/6
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